Алгебра и нач анализа_Реш экз зад 11кл из Сборн заданий для экз_Дорофеев_Решения (991497), страница 20
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log5(x – 8)2 = 2 + 2log5(x – 2); log5(x–8)2 = log525 + log5(x – 2)2;2222(х – 8) = 25 ⋅ (х – 2) ; х – 16х + 64 = 25х – 100х + 100;24х2 – 84х + 36 = 0; 2х2 – 7х + 3 = 0; D = 25; О.Д.З. х ≠ 8; х > 21Ответ: 3.х1 = 3, x2 = . – не подходит в О.Д.З.216.3. log 9 x2 ( 6 + 2 x − x 2 ) = ;2⎧⎧ x 2 − 2 x − 6 < 0,⎪6 + 2 x − x 2 > 0,1⎪ 2⎪⎪⎨9 x ≠ 1,⎨x ≠ ± ,31⎪⎪2⎪⎩6 + 2 x − x 2 = ( 9 x 2 ) 2 ; ⎪⎩6 + 2 x − x = 3 x .26 + 2х – х = 3|x|.1) х ≥ 0; 6 + 2х – х2 = 3х; х2 + х – 6 = 0; х1 = -3, х2 = 2. Значит, х = 2.2) х < 0; 6 + 2х – х2=-3х; х2–5х – 6 = 0; х1 = 6, х2 = -1.
Значит, х = -1.Ответ: 2; -1.6.4. logx-3(x2 – 4x)2 = 4; 2logx-3|x2 – 4x| = 4; logx-3|x2 – 4x| = 2;⎧ x − 3 > 0,⎧ x > 3,⎪⎪⎨ x − 3 ≠ 1,⎨ x ≠ 4,⎪ x 2 − 4 x = ( x − 3 )2 ; ⎪ x 2 − 4 x = ( x − 3 )2 ;⎩⎩1) x3 − 4 х > 0 ⇔ x ∈ ( −∞;0 ) ∪ ( 4; ∞ ) .
х2 – 4х = х2 – 6х + 9; х = 4,5;2) x 2 − 4 х < 0 ⇔ x ∈ ( 0;4 ) . 4х – х2 = х2 – 6х + 9;2042х2 – 10х + 9 = 0; х =5± 7 5− 75+ 7< 3 . х1 = 4,5; x2 =;.2225+ 7.2x6.5. log3(3 – 8) = 2 – x;⎧3x − 8 > 0,3х – 8 = 32-х, 32х – 8 ⋅ 3х – 9 = 0;⎨ x2− x⎩3 − 8 = 3 ;Ответ: x1 = 4,5; x2 =сделаем замену 3х = t, t > 0: t2 – 8t – 9 = 0; t1 = 9, t2 = -1.Ответ: 2.t > 0 ⇒ 3х = 9; 3х = 32; х = 2.6.6. log7(7-x + 6) = 1 + x; 7-x + 6 = 71+x; 1 + 6 ⋅ 7x – 7 ⋅ 72x = 0, замена17х = t, t > 0; 7t2 – 6t – 1 = 0; t1 = 1, t2 = − ; t = 1;77х = 1; 7х = 70; х = 0.Ответ: 0.6.7. log2(2x – 7) = 3 – x; 2х – 7 = 23-х; 22х – 7 ⋅ 2х – 8 = 0; сделаемзамену 2х = t, t > 0; t2 – 7t – 8 = 0; t1 = 8, t2 = -1; 2х=8; 2х = 23; х = 3.Ответ: 3.6.8.
log4(4-x + 3) = x + 1. 4-x + 3 = 4x+1; 1 + 3 ⋅ 4x = 4 ⋅ 42x;Пусть t = 4x, t > 0; 4t2–3t – 1 = 0; t1 = -1/4 < 0; t2 = 1; 4x = 1 ⇔ x = 0Ответ: х = 0.6.9. log6(6-x + 5) = 1 + x; 6-x + 5 = 61+x; 6 ⋅ 62x – 5 ⋅ 6x – 1 = 0,1пусть 6х = t, t > 0; 6t2 – 5t – 1 = 0; t1 = 1, t2 = − ; 6х=1; 6х=60; х = 0.6Ответ: 0.6.10. log5(5x – 4) = 1 – x; 5х – 4 = 51-х; 52х – 4 ⋅ 5x – 5 = 0; 5x = t, t > 0;t2 – 4t – 5 = 0; t1 = 5, t2 = -1; 5x = 5; x = 1.Ответ: 1.6.11. 2log7(x – 2) = -2 + log7(x – 10)2;22⎞⎛ 1log 7 ( x − 2 ) = log 7 ⎜ ( x − 10 ) ⎟ , x > 2, x ≠ 10;⎝ 49⎠49(х – 2)2 = (х – 10)2; 49х2 – 196х + 196 = х2 – 20х + 100;248х2 – 176х + 96 = 0; 3х2 – 11х + 6 = 0; D = 49; x1 = , х2 = 3.3Ответ: 3.16.12.
log ( x−6 )2 ( x 2 − 5 x + 9 ) = ; х2 – 5х + 9 = |x – 6|, (х – 6)2 ≠ 0,2(х – 6)2 ≠ 1. Значит, х ≠ 6, х ≠ 7, х ≠ 5.21) х > 6; х – 5х + 9 = х – 6; х2 – 6х + 15 = 0;205D= −6 < 0 , корней нет;42) x<6; х2–5х + 9 = -х + 6; х2 – 4х + 3 = 0; х1 = 1, х2 = 3. Ответ: 1; 3.6.13. (2х2 – 5х + 2)(log2x18x + 1) = 0; О.Д.З.
x > 0; x ≠ 1/2.1) 2x2 – 5x + 2 = 0; x1 = ½; x2 = 2. Подставляя в О.Д.З имеем: х = 2.11112) log 2 x 18 x + 1 = 0;18 x = ; x 2 = ; x = ± . Ответ: 2; .2x3666⎛⎞⎛⎞6.14. ( x2 − 7x + 10) ⎜⎜ log x 8x + 1⎟⎟ = 0; ( x 2 − 7 x + 10 ) ⎜⎜ log x 16 + 2 ⎟⎟ = 0;⎝ 2⎠⎝ 2⎠⎡ ⎧ x 2 − 7 x + 10 = 0,⎢ ⎪ x > 0,⎢⎨⎢ ⎪⎩ x ≠ 2,⎢log x 16 = −2;⎢⎣ 2⎧⎪ x > 0,⎪х = 5 или ⎨ x ≠ 2,⎪ 2 1⎪x = ;⎩4⎧⎪⎪⎪ x > 0,или ⎨ x ≠ 2,⎪⎛ x ⎞−2⎪⎜ ⎟ = 16;⎩⎪⎝ 2 ⎠⎧ ⎡ x = 5,⎪⎢ x = 2,⎪⎣⎨ x > 0,⎪x ≠ 2⎪⎩1х = 5 или x = .2Ответ: 5;1.26.15. (2 x − 3) 3x 2 − 5 x − 2 = 0 ,3⎧⎪x = 2 ,1⎪или х = 2, или x = − ;⎨ ⎡ x ≥ 2,3⎪⎢1⎪⎢ x ≤ − ,3⎣⎩Ответ: 2; −1/ 3 .х = 2 или x = −1/ 3 .⎡ ⎧2 x − 3 = 0,⎢ ⎨3x 2 − 5 x − 2 ≥ 0,⎢⎩ 2⎢⎣3x − 5 x − 2 = 0;6.16.
(2 x 2 − 3 x − 2) 3x + 1 = 0.⎧2 x 2 − 3x − 2 = 0,1) ⎨⎩3 x + 1 ≥ 0или12) x = − .31) 2х2 – 3х – 2 = 0;D = 25;20611х1 = 2, x2 = − < − ;232) 3х + 1 = 0.Ответ: 2; −1.36.17. ( 6 x − 5 ) 2 x 2 − 5 x + 2 = 0.⎧6 x − 5 = 0,1) ⎨ 2⎩2 x − 5 x + 2 ≥ 0;5⎧⎪⎪ x = 6 ,⎨ ⎛1⎪2 ⎜ x − ⎟⎞ ( x − 2 ) ≥ 0;2⎠⎪⎩ ⎝5⎧⎪x = 6 ,⎪⎨⎡ x ≤ 1 ,⎪⎢2⎪⎢⎩ ⎣ x ≥ 2.Система решений не имеет.2) 2х2 – 5х + 2 = 0; D = 9; x1 =1, х2 = 2.2Ответ: 2;1.26.18.
(3x 2 − x − 2) 2 x − 1 = 0.⎡ ⎧3 x 2 − x − 2 = 0,⎢ ⎨ 2 x − 1 ≥ 0,⎢⎩⎢⎣ 2 x − 1 = 0;23х2 – х – 2 = 0; D = 25; х1 = 1, x2 = − .3⎧ ⎡ x = 1,⎪⎢211⎪⎢ x = − ,⎨⎣3 или x = ; х = 1 или x = ;22⎪1⎪x ≥⎩2Ответ: 1;1.26.19. (7 x + 2) 4 x − 3x 2 − 1 = 0.⎡ ⎧7 x + 2 = 0,⎢ ⎨ 4 x − 3 x 2 − 1 ≥ 0, 4х – 3х2 – 1 = 0; 3х2 – 4х + 1 = 0; х = 1, x = 1 ;12⎢⎩3⎢⎣ 4 x − 3 x 2 − 1 = 0;⎡⎧2⎢ ⎪⎪ x = − 7 ,⎢⎨⎢ ⎪3 ( x − 1) ⎛ x − 1 ⎞ ≤ 0,⎜⎟⎢ ⎪⎩3⎠⎝⎢ x = 1,⎢⎢x = 1;⎢3⎢⎢⎢⎣Ответ: 1;⎡⎧2⎢ ⎪⎪ x = − 7 ,⎢⎨⎢ ⎪ x ∈ ⎡ 1 ;1⎤ ,⎢ ⎪⎩ ⎣⎢ 3 ⎦⎥⎢ x = 1,⎢⎢x = 1;⎢3⎢⎢⎢⎣⎡ x = 1,⎢1⎢x = .3⎣1.32076.20.
(3x − x 2 − 2) 7 x + 4 = 0;⎡ ⎧3 x − x 2 − 2 = 0,⎢ ⎨7 x + 4 ≥ 0,3х – х2 – 2 = 0; х2 – 3х + 2 = 0; х1 = 1, х2 = 2;⎢⎩⎢⎣7 x + 4 = 0;⎡ ⎧ ⎡ x = 1,⎢ ⎪⎪ ⎢ x = 2,⎢⎨⎣⎢⎪ x ≥ − 4 ,⎢ ⎪⎩7⎢4⎢x = − ;⎢⎣7⎡⎢ x = 1,⎢ x = 2,⎢⎢x = − 4.⎢⎣7Ответ: 1; 2; −4.76.21. (3x + 4) −3x − 2 x 2 − 1 = 0;⎧3 x + 4 = 0,⎨2⎩−3 x − 2 x − 1 ≥ 0или –3х – 2х2 – 1 = 0;-3х – 2х2 – 1 = 0;12х2 + 3х + 1 = 0; D = 1; х1 = -1, x2 = − ;24⎧⎪⎪ x = − 3 ,⎨⎪2 ( x + 1) ⎛⎜ x +⎪⎩⎝6.22.1⎞⎟ ≤ 0;2⎠4⎧⎪⎪ x = − 3 ,⎨ ⎡1⎪ x ∈ ⎢ −1; − ⎤⎥ ,2⎦⎪⎩ ⎣Ответ: -1; −1.2(4 x − x 2 − 3) 5 x − 8 = 0;⎧ x 2 − 4 x + 3 = 0,8⎪⎧4 x − x 2 − 3 = 0,или 5х – 8 = 0; ⎨или x = ;8⎨5⎩5 x − 8 ≥ 0⎪⎩ x ≥ 5⎧ ⎡ x = 3,⎪⎪ ⎢ x = 1,88⎣или x = ; х = 3 или x = .⎨55⎪x ≥ 8⎪⎩52Ответ: 3; 1,6.xx⎞⎛6.23.
1 + sin 3 x = ⎜ cos − sin ⎟ ;22⎠⎝xxxx1 + sin 3 x = cos 2 − 2sin cos + sin 2 ;2222sin3x + sin x = 0; 2cos x ⋅ sin2x = 0; cos x = 0 или sin2x = 0;208x=π2+ π n, n ∈ Z ;Ответ: x =2x = πk; x =π2k, k ∈ Z.πm, m ∈ Z .26.24. 2sin 2x = (cos x + sin x)2; 2sin22x – sin2x – 1 = 0;1πsin2x = 1 или sin 2 x = − ; 2 x = + 2π n, n ∈ Z ;22ππk +1 πk +1 π+ k , k ∈ Z.2 x = ( −1)+ π k , k ∈ Z ; x = + π n, n ∈ Z; x = ( −1)412 26ππk +1 πОтвет: (1 + 4n ) ; ( −1)+ k , n, k ∈ Z.412 26.25.
cos9x – cos7x + cos3x – cos x = 0;(cos9x–cos x)–(cos7x–cos3x) = 0; -2sin5x ⋅ sin4x + 2sin5x ⋅ sin2x = 0;sin5x(sin4x – sin2x) = 0;sin5x = 0 илиsin4x – sin2x = 0;2cos3x sin x = 0;5x = πm, m ∈ Z;ππx = m, m ∈ Z ;cos3x = 0, x = (1 + 2n ) , n ∈ Z ;56или sin x = 0, x = πk, k ∈ Z.ππОтвет: m;(1 + 2n ) , n, m ∈ Z.566.26. cos7x+sin8x=cos3x–sin2x; (cos7x–cos3x) + (sin8x + sin2x) = 0;-2sin5x sin2x + 2sin5x cos3x = 0; sin5x(sin2x – cos3x) = 0;sin5x = 0илиsin2x – cos3x = 0;25x = πm, m ∈ Z;x=π5m, m ∈ Z ;⎛π⎞sin 2 x − sin ⎜ − 3x ⎟ = 0;⎝2⎠⎛ π x ⎞ ⎛ 5x π ⎞2cos ⎜ − ⎟ sin ⎜ − ⎟ = 0;⎝ 4 2⎠ ⎝ 2 4 ⎠3⎛π x ⎞1) cos ⎜ − ⎟ = 0; x = π + 2π k , k ∈ Z ;2⎝ 4 2⎠π 2⎛ 5x π ⎞2) sin ⎜ − ⎟ = 0; x = + π n, n ∈ Z .10 5⎝ 2 4⎠ππ 23π+ π n;+ 2π k , m, n, k ∈ Z.Ответ: m;510 522096.27. sin x–sin2x+sin5x+sin8x=0; (sin x + sin5x) + (sin8x – sin2x)=0;2sin3x cos2x + 2sin3x cos5x = 0; sin3x(cos2x + cos5x) = 0;sin3x = 0илиcos2x + cos5x = 0;73πx = m, m ∈ Z ;2cos x cos x = 0;3x = πm, m ∈ Z;32277ππ 2x = + π k; x = + π k , k ∈ Z ;1) cos x = 0;2227 733ππ 22) cos x = 0;x = + π n; x = + π n, n ∈ Z .2223 3ππm;(1 + 2k ) , m, k ∈ Z.376.28.
sin x+sin3x–sin5x–sin7x=0; sin x + sin3x – (sin5x + sin7x) = 0;cos x(sin2x – sin6x) = 0;cos x = 0илиsin2x – sin6x = 0;Ответ:x=π2+ π m, m ∈ Z ;-2cos4x sin2x = 0;cos4x = 0; x =π8(1 + 2n ) , n ∈ Z ;или sin2x = 0; x =ππ2k, k ∈ Z.π(1 + 2n ) ; k , где k, n ∈ Z.826.29. cos2x + cos6x + 2sin2x = 1; cos2x + cos6x = 1 – 2sin2x;Ответ:cos2x+cos6x=cos2x; cos6x=0; 6 x =Ответ:π12π2+ π m; x =π12(1 + 2m ) , m ∈ Z.(1 + 2m ) , m ∈ Z.1= 0;sin x cos x22sin 2x + 2 = 0; 1 – cos4x = -2; cos4x = 3 – нет решений,т.к. |cos α| ≤ 1.Ответ: нет решений.6.31. cos x + cos2x + cos3x = 0;(cos x + cos3x) + cos2x = 0;2cos2x cos x + cos2x = 0; cos2x(2cos x + 1) = 0;cos2x = 0или2cos x + 1 = 0;π1cos x = − ;2 x = + π m, m ∈ Z ;226.30.
4cos x ⋅ sin x + (tg x + ctg x) = 0; 2sin 2 x +210x=π4(1 + 2m ) , m ∈ Z;2x = ± π + 2π n, n ∈ Z .3π2(1 + 2m ) ; ± π + 2π n, m, n ∈ Z.436.32. sin x + sin3x = 4cos2x;2sin2x cos x – 4cos2x = 0; 4cos2x(sin x – 1) = 0;Ответ:cos x = 0; x =π2+ π k, k ∈ Z;ππ+ 2π m, m ∈ Z . Ответ: (1 + 2k ) , k ∈ Z.226.33. cos x = cos3x + 2sin2x; cos3x – cos x + 2sin2x = 0;-2sin2x sin x + 2sin2x = 0; 2sin2x(sin x – 1) = 0;или sin x = 1; x =sin2x = 0; 2x = πm, m ∈ Z; x =π2m, m ∈ Z .ππ+ 2π k , k ∈ Z ; Ответ: l , l ∈ Z.226.34. 8sin22x + 4sin24x = 5; 4(1 – cos4x) + 4sin24x = 5;4cos24x + 4cos4x – 3 = 0. Пусть cos4x = y, тогдаD−2 − 4−2 + 4 1= 4 + 12 = 42 ; y1 == −1,5, y2 == ;4у2+4у–3=0;44421) cos4x = -1,5 – решений нет, т.к.
|cos x| ≤ 1;1ππ π2) cos 4 x = ; 4 x = ± + 2π k ; x = ± + m , где m ∈ Z.2312 2или sin x = 1; x =Ответ: ±ππ+ m, m ∈ Z .12 26.35. sin23x + sin24x = sin25x + sin26x; 1 – cos6x + 1 – cos8x == 1 – cos10x + 1 – cos12x; cos12x – cos6x = cos8x – cos10x;-2sin9x sin3x = 2sin9x sin x; sin9x(sin3x + sin x) = 0;sin9x = 0; 9x = πm, m ∈ Z; x =π9m, m ∈ Z ;или sin3x+sin x=0; 2sin2x cos x=0; sin2x=0; 2x = πk; x =cos x = 0; x =Ответ:π9m;π2π2+ π n; x =π2π2k,k ∈ Z;(1 + 2n ) , n ∈ Z.l , m, l ∈ Z.2116.36.
sin2x + sin22x + sin23x + sin24x = 2; 1 – cos2x + 1 – cos4x ++ 1 – cos6x + 1 – cos8x = 4; (cos2x + cos8x) + (cos4x + cos6x) = 0;2cos5x cos3x + 2cos5x cos x = 0; cos5x(cos3x + cos x) = 0;cos5x = 0илиcos3x + cos x = 0;5x =x=π2π+ π m, m ∈ Z ;2cos2x cos x = 0;(1 + 2m ) , m ∈ Z ;101) cos2x = 0; 2 x =x=π4+π2π2+ π k, k ∈ Z;k, k ∈ Z;2) cos x = 0; x =πππ2+ π n, n ∈ Z .π(1 + 2m ) ; (1 + 2k ) ; (1 + 2n ) , k, m, n ∈ Z.104226.37.
cos 3x+cos24x+cos25x=1,5; 1+cos6x+1+cos8x + 1 + cos10x = 3;(cos6x+cos10x)+cos8x=0; 2cos8xcos2x+cos8x=0; cos8x(2cos2x+1)=0;cos8x = 0;или2cos2x = -1;π1cos 2 x = − ;8 x = + π n, n ∈ Z ;22π π2ππx = + n, n ∈ Z ; 2 x = ±+ 2π k , k ∈ Z ; x = ± + π k , k ∈ Z .316 83Ответ:πππ+ n; ± + π k , n, k ∈ Z.16 8326.38. cos x + cos22x = cos23x + cos24x; 1 + cos2x + 1 + cos4x == 1 + cos6x + 1 + cos8x; cos2x + cos4x = cos6x + cos8x;2cos3x cos x = 2cos7x cos x; cos x(cos3x – cos7x) = 0;cos x = 0илиcos3x – cos7x = 0;Ответ:x=π2+ π m, m ∈ Z ;2sin5x sin2x = 0;1) sin5x = 0; 5x = πk; x =2) sin2x = 0; 2x = πn; x =Ответ:212π2l;π5m , где l, m ∈ Z.π5π2k, k ∈ Z;n, n ∈ Z .6.39. 2cos24x – 6cos22x + 1 = 0;2cos24x – 3(1 + cos4x) + 1 = 0; 2cos24x – 3cos4x – 2 = 0;3−513+5cos 4 x ==−cos 4 x =или=2 2⋅222⋅22решений нет, т.к. |cos α| ≤ 1;4 x = ± π + 2π n, n ∈ Z ;3πx=±πππ+ n, n ∈ Z .Ответ: ± + n, n ∈ Z .6 26 26.40.
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